A circle has the center of A and the radius of 5. |BC| = 8; |AC|=|AB|=5
Find the area of the gray bit of the circle.
Use the cos rule to find angle A (I will use radians because it's easier)
$\displaystyle \cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{5^2+5^2-8^2}{2 \cdot 5 \cdot 5} = -\frac{7}{25}$
$\displaystyle A = \arccos \left(-\frac{7}{25}\right) \approx 1.855$
The area of a sector is given by $\displaystyle A = \frac{1}{2}r^2 \theta$. In this case $\displaystyle \theta$ is angle A and $\displaystyle r=5$
$\displaystyle A_s = \frac{1}{2} \cdot 5^2 \cdot \arccos \left(-\frac{7}{25}\right)$
Since this includes the white triangle we must remove it's area which is given by $\displaystyle A_t = \frac{1}{2} bc \sin A = \frac{1}{2} \cdot 5^2 = \frac{25}{2}$
$\displaystyle A_g = A_s - A_t = \frac{1}{2} \cdot 5^2 \cdot \arccos \left(-\frac{7}{25}\right) - \frac{25}{2}$
Evaluate for the answer, I get an answer of $\displaystyle 10.7 \text{ 3sf }$